Module-3: CALCULUS

 Reduction formulae (without proof)

 Evaluation of definite and improper integrals.
 Beta and Gamma functions and their properties.
 Tracing of standard curves: Strophoid, Leminscate, cardioid, and Astroid.
 Applications of definite integrals to evaluate surface areas and volumes of revolutions

Introduction

Reduction formulae are basically a recurrence relation which reduces integral of functions of
higher degree in the form   f  x  dx ,   f x  g x  dx
n m n
(where m and n are non-negative integrals) to
lower degree. The successive application of the recurrence relation finally ends up with a function of
degree 0 or 1 so that we can easily complete the integration process.

We discuss three standard reduction formulae in the form of definite integrals & the
evaluation of them with standard limits of integration.

n n n n n n
Reduction formulae for integration of sin x , cos x , tan x , cot x , sec x and cos ec x .
(Without derivations, where ' n ' being positive integer)

Definition:

A formula which expresses (or reduces) the integral of the nth indexed function in terms of
that of n  1 th indexed or lower indexed function is called a reduction formula.

The two results are very useful in the evaluation of reduction formulae.

Integration by parts:

If u and v are functions of x then  uvdx  u vdx     vdx udx , where u  represents
differentiation of the function u .

Bernoulli's Formula:

If u and v are functions of x then  uvdx  uv 1  u v2  u v3  u v4  ....

where, u  , u and u  …are derivatives of the function

u , v1, v2 and v3 … are integrals of the function v .

  sin
n
Reduction formula for x dx where n is a positive integer

Consider I n   sin n x dx which can also be written as

I n   sin ( n 1) x  sin x dx … (1)

Integrating by parts (1) we get

In  sinn1 x( cos x)   ( cos x)(n 1)sinn2 x cos xdx

In   sin ( n 1) x cos x  (n  1)  sin ( n2) x cos2 x dx
In   sin ( n 1) x cos x  (n  1)  sin ( n 2) x(1  sin 2 x) dx
I n   sin ( n 1) x cos x  ( n  1)  (sin ( n  2 ) x  sin n x ) dx

I n   sin ( n 1) x cos x  ( n  1)  sin ( n 2) xdx  (n  1)  sin n xdx

I n  (n  1) I n   sin ( n1) x cos x  (n  1) I n 2
nI n   sin (n1) x cos x  (n  1) I n2
 sin( n1) x cos x (n  1)
In   I n 2
n n
Where I n   sin n x dx and I n  2   sin n  2 x d x

Applying the limits of integration i.e., from 0 to  for the above reduction formula we get
2

 
2
 sin n 1 x cos x 2 (n  1)
I n   sin x n
 I n2
0
n 0
n

n 1
In  I n2
n

n 1
Consider In  I n2 … (2)
n

Replace n by  n  2  in equation (2)

n -3
I n -2  I n -4 … (3)
n-2

Replace n by ( n  2) in equation (3)

n-5
I n -4  I n -6
n-4

We finally have
  n -1 n - 3 n - 5 1
 n n - 2 n - 4 ... 2 I 0 n is even
In  
 n -1 n - 3 n - 5 .. 2 I n is odd
 n n - 2 n - 4 3 1

 /2
  /2
Where, I 0  
0
sin 0 x dx 
2
and I1   sin x dx  1
0

 n -1 n - 3 n - 5 1 
 n n - 2 n - 4 ... 2 2 n is even
In  
 n -1 n - 3 n - 5 ... 2 1 n is odd
 n n - 2 n - 4 3

 cos
n
Reduction formula for x dx where n is a positive integer

Consider I n   cos n x d x … (1)

I n   cos n 1 x cos x dx

Integrating the above equation by parts we get

I n  cos n 1 x sin x   (sin x )( n  1) cos n  2 x (  sin x ) dx

 cos n 1 x sin x  (n  1)  cos n 2 x sin 2 xdx

 cos n 1 x sin x  (n  1)  cos n 2 x(1  cos 2 x)dx
 cos n 1 x sin x  (n  1)  cos n 2 xdx  (n  1)  cos n xdx
 cos n 1 x sin x  (n  1) I n  2  (n  1) I n

I n  (n  1) I n  cosn1 x sin x  (n  1) I n 2

nI n  cosn1 x sin x  (n 1) I n2

cos n 1 x sin x ( n  1)
In   I n2 … (2)
n n


2
Let I n =  cos n x dx
0
From (2),
 

 cos n  1 x sin x 
2

n 1
In    I n2
 n  n
  0

 
But cos   0  sin 0
2

n 1
Thus I n  I n2
n
 /2  /2  /2
 
 cos n x dx   cos n   x  dx   sin
n
Note : xdx
0 0 2  0

 n 1 n  3 n  5 2
 . . ................. , If is n odd

 n n2 n4 3
 cos n x  
2

 n  1 . n  3 . n  5 ............ 1 .  If is n even
0

 n n  2 n  4 2 2

Problems

1. Evaluate  sin 5 x dx
 sin 4 x cos x 4
Solution:  sin x dx    sin 3 xdx
5

5 5

 sin 4 x cos x 4 8
 sin xdx   sin 2 x cos x  cos x  c
5

5 15 15
 /2

 sin
7
2. Evaluate x dx
0
 /2
6 4 2
 sin x dx    1
7
Solution:
0
7 5 3

 /2
16
 sin x dx 
7

0
35

 /6

 sin
6
3. Evaluate 3x dx
0
 /6
I  sin
6
Solution: 3 x dx
0

1
Put 3x  t , dx  dt , when x  0 , t  0
3

 
when x  ,t
6 2
  /2
1
I  sin t  dt
6

0
3

 /2
1 1 5 3 1 
I  sin tdt     
6

3 0
3 6 4 2 2

5
I
96

1
4. Evaluate  (1  x )
0
2 n
dx

1
Solution: I dx
0
(1  x 2 n
)


Put x  tan , dx  sec  d
2
when x  0   0 ; when x    
2
 /2
1
I 
0
(1  tan  )
2 n
sec 2  d

 /2

 cos  d
2 n 2
I
0


1 2n  3 2 n  5 2 n  7 1 
 (1  x
0
2 n
)
dx    ... 
2n  2 2n  4 2n  6 2 2

3
x3
5. Evaluate 
0
3 x
dx

3
x3
Solution: I  dx
0
3 x


Put x  3sin  , dx  6 sin cos d x  0   0 ; when
2
when x3 
2

 /2
27sin6 
I 
0
3  3sin 2 
6sin  cos d

 /2
3 1 
I  18  sin 4  d  18   
0
4 2 2
 27
3
x3

0
3 x
dx 
8


x2
6. Evaluate 
0 (1  x6 )7
dx


x2
Solution: I  dx
0 (1  x6 )7


Put x  tan , 3x dx  sec  d when x  0   0 ; when x    
3 2 2

2
 /2
1 1
I  sec  sec  d
2
7
3 0

 /2
1 1 4 2
I  cos  d  3  5  3 1
5

3 0


x2 8
 dx 
0 1  x  6 7 135


7. Evaluate  0
x sin 7 xdx
Solution:

I   x sin 7 xdx
0

I   (  x) sin 7 xdx
0
 
I    sin 7 xdx   x sin 7 xdx
0 0

I    sin 7 xdx  I
0

2 I    sin 7 xdx
0
2a a
[use  f ( x)dx  2 f ( x)dx if f (2a  x)  f ( x)]
0 0

2 I  2  2 sin 7 xdx
0

I    2 sin 7 xdx
0

6 4 2 16
I  1
753 35
 
8. Evaluate  0
4 sin 5  cos 5 d

Solution:


I   4 sin 5  cos 5 d
0

I   4 (sin  cos  ) 5 d
0
 5
2 
I  4
 sin  cos   d
0
2 

1 4 5
32 0
I sin 2d

Put 2  t and d  dt
2
 0 t 0
 
  t 
4 2

1 2 5
64 0
I sin tdt

1 42 1
I 1
64 5 3 120

 1  cos
9. Evaluate 
0 1  cos
sin 2 d

Solution:
  1  cos 
0 1  cos 
sin 2 d

 2 sin 2 
2 (2 sin  cos  ) 2 d
0
2 cos  2 2 2
2

 4 sin 4  d
0 2
Put    and d  2d
2
If   0    0

   
2


 4 sin 4 2d
2
0

3 1  24
8 
422 6

 x7
10. Evaluate 
0
(a 2  x 2 )
dx

Solution:

 x7
0
(a 2  x 2 )
dx

Put x  a sin  and dx  a cos d

If x  0    0

x  a  
2

a sin  7 7

0
2

(a  a 2 sin 2  )
2
a cos d


a 7  2 sin 7d
0

642 16
a7 1  a7
753 35
  x2
11. Evaluate 0
(1  x 2 ) 2
dx

Solution:

 x2

0
(1  x 2 ) 2
dx

Put x  tan  and dx  sec 2 d

If x  0    0

x    
2

tan 2


0
2

(1  tan  ) 2 2
sec 2 d


0
2
sin 2 d
1 

22 4

1 x10
12. Evaluate 0
(1  x 2 )
dx

Solution:

1 x10

0
(1  x 2 )
dx

Put x  sin  and dx  cos d

If x  0    0

x 1 
2

sin  10


0
2

(1  sin 2  )
cos d


0
2 sin 10 d
9 7 5 3 1  63

10 8 6 4 2 2 512
 2 x4
13. Evaluate 
0
(4  x 2 )
dx

Solution:

2 x4
0
(4  x 2 )
dx

Put x  2 sin  and dx  2 cos d

If x  0    0

x  2  
2

2 sin 
4 4

0
2

(4  2 2 sin 2  )
2 cos d


16 2 sin 4 d
0

31
16  3
422

Evaluation of sin m x cos n x integrals:

Let I m ,n   sin m x cos n xdx

I m ,n   sin m 1 x cos n x sin x dx

 Note : 
 m 1 
 p  sin xcox x
n

 
dp  (m  1) sin m 2 x cos x cos n x  sin m1 x(n cos n1 x( sin x))
 
dp  (m  1) sin m 2 x cos n1 x  n sin m x cos n1 x 

Integrating the above equation by parts we get

 I m,n  sin m1 x cos n x  sin xdx   (sin xdx)[(m  1) sin m 2 x cos n1 x  n sin m x cos n 1 x]dx
I m,n   sin m1 x cos n1 x   cos x[(m  1) sin m 2 x cos n1 x  n sin m x cos n 1 x]
I m,n   sin m1 x cos n1 x  (m  1)  cos n 2 x sin m 2 xdx  n  sin m x cos n xdx
I m,n   sin m1 x cos n1 x  (m  1)  cos n 2 x sin m 2 xdx  nI m,n
I m,n [1  n]   sin m1 x cos n1 x  (m  1)  cos n x[1  sin 2 x] sin m 2 xdx
I m,n [1  n]   sin m1 x cos n1 x  (m  1)  cos n x sin m 2 xdx  (m  1)  cos n x sin m xdx
sin m1 x cos n1 x (m  1)
I m,n    I m  2, n
( m  n) mn

Applying limits to the above equation from 0 to , we get
2

 /2  /2
 sin m1 x cosn1 x m 1
I m,n   sin x cos x dx  
m n
I m2,n
0
mn 0
mn

m 1
I m,n  I m2,n … (1)
mn
which is the reduction formula by reducing the power of sin x function.

Note: If we reduce the power of cos x function then we will get the reduction formula as

n 1
I m,n  I m,n 2
mn

Replace m by  m  2  in equation (1) we get

m3
I m  2, n  I m  4, n … (2)
mn2

Replace m by  m  4  in equation (1) we get

m5
I m  4, n  I m  6, n … (3)
mn4

If m is even then by putting m  2 in equation If m is odd then by putting m  3 in equation (1)

(1)
2
I 3, n  I1, n
1 3  n
I 2, n  I 0, n
2n
 /2  /2
 cosn1
I1, n   sin x cos xdx 
n

0
n 1 0
  /2
1

I 0, n   cos
n
x dx
n 1
0

 n 1 n  3 1 
 n  n  2 ... 2  2 n is even

 n  1  n  3 ... 2 1 n is odd
 n n2 3

Summarizing the above cases in (1), we get

 n 1 n  3 1 
 m 1  ...  if m is even n is even
m3 m 5 1  n n2 2 2
   ... 
m 1 m  n  2 m  n  4 2  n n 1 n  3 2
I m ,n   ...  1 if m is even n is odd
  n n2 3
 m 1 m3 m5 2 1 if m odd n is odd
   ...  , if m is odd n is even
m  n m  n  2 m  n  4 3  n n 1

Evaluate the following integrals:

 /6

 cos
4
1. Evaluate 3 x sin 3 6 x dx
0
 /6
I  cos
4
Solution: 3 x sin 3 6 xdx
0

Put 3x  t , we get when x  0 t 0

 
when x t
6 2
 /2
I  cos
4
t sin 3 2t (1 / 3 dt )
0

 /2
1
I  (cos
4
t ) (2sin t cos t )3 dt
3 0

 /2
8 8 2 1
I  sin t cos 7 tdt  . .
3

3 0
3 10 8
 6
1
 cos
4
3 x sin 3 6 x dx 
0
15
 


2. Evaluate (sin x) (1  cos x ) dx
2 3

0

Solution: I   (sin 2 x ) (1  cos x )3 dx
0

Put x  2t , we get when x  0 t 0

when x  t  
2
 /2
I  2  (sin 2 2t ) (1  cos 2t ) 3 dt
0

 /2
I  64  sin
8
t cos 2 tdt
0

7  5  3 1 1  7
I  64   
10  8  6  4 2 2 8

x5
3. Evaluate 0 (1  x 2 )6 dx
Solution: Put x  tan  when x  0  0


when x 
2
  /2  /2
x5 tan 5  4.2 1
0 (1  x2 )6 dx   sec 2  d   sin  cos5  d  
5

0
sec 
12
0
10.8.6 60


x6
4. Evaluate  dx
0
(1  x 2 )9/2
Solution: Put x  tan  when x  0  0


when x 
2
  /2  /2
x6 tan 6  1
0 (1  x2 )9/ 2 dx   sec 2  d   sin  cos  d 
6

0
sec9  0
7
 2a
5. If n is a positive integer, show that  x n 2 ax  x 2 dx  a 2   .
a 2 n  1! ,
n

0   n  2 ! n!
2 
2a

x 2 ax  x 2 dx
3
hence evaluate
0

2a
I x 2ax  x 2 dx
n
Solution:
0

Put x  2a sin 2  , dx  4a sin  cos  d ,  varies from 0 to 

2

 /2
n n
I 2 a sin 2n  4a 2 sin 2   4a 2 sin 4  4a sin  cos  d
0
 /2

 2sin  cos 2  d
n2 2n2
I  (2a)
0

(2n  1)(2n  1)...1 

I  (2a)n2  2  .
(2n  4)(2n  2)...2 2

(2n  1)(2n 1)...1

I  (2a)n2 
2n2 (n  2)(n  1)...1

(2n  1) (2n  1) (2n  3) 1 1 

I  ( 2a ) n  2 2   ....  
(2n  4) (2n  2) ( 2 n) 4 2 2
(2n  1) (2n  1) (2n  3) 1 1
I  ( 2a ) n  2    .... 
(2n  4) (2n  2) ( 2n ) 4 2
(2n  1)2n(2n  1)(2n  2)(2n  3)....1
I  ( 2a ) n  2 
2(n  2)2(n  1)2(n).....2(2)2(1)[2n  2(n  1).....2(1)]
(2n  1)!
I  2 n  2 a n  2 n  2
2 (n  2)!(2 n )(n!)
(2n  1)!
I  a n  2 n
(2 )(n!)(n  2)!
n
 a  (2n  1)!
I  a   2

 2  (n!)(n  2)!
Next, for n  3 ,
 2a
a 3 2 ( 6  1)!
 x 2 ax  x dx  
3 2

0
2 3  3!(5!)
a 5 7!
 
8  6  5!
(7  6!)a 5

8  6!
7a 5

8

Beta, Gamma functions and properties of Beta and Gamma functions:

Beta Function:


The integral  (m, n)  x (1  x) dx
m 1 n 1
(m,n  0) is called a Beta function.
0

Gamma Function


n 1  x
The integral (n)  x e dx (n  0) is called a gamma function (n ) .
0

Properties

Property 1 : (1)  1

Proof : By definition


 (1)   x 0 e  x dx   e  x  0  1
0

(1)  1

Property 2 : Recurrence formula (n  1)  n(n)

Proof : By definition

 

e x  e x n 1 
  
n
 ( n  1)   x e dx
n x
x  nx dx  0  n  x n 1e  x dx
0
1  0 0 1 0

(n  1)  n(n)
 Case (i) If n is a positive integer

Proof:

(n  1)  (n)
(n)  (n  1)(n  1)
(n  1)  (n  2)(n  2)
 
(3)  2(2)
(2)  1(1)

Now we have
(n  1)  n[(n  1)(n  2)(n  3)....2  1](1)
(n  1)  n!(1)
(n  1)  n!

  n 1
Case (ii) If n is a negative real number then   n  .
n
Case (iii) If n  0 or n is negative integral value then (n) is not defined.

Properties:

Property 1 : Show that  ( m, n )   ( n, m )

1
Proof :  (m, n)   x m 1 (1  x ) n1 dx
0

1
m 1
1  (1  x)
n 1
  (1  x) dx ¸
0
a a
by using  f ( x) dx   f (a  x) dx
0 0

1
  (1  x ) m 1 x n 1 dx
0

 (m, n)   (n, m)
  /2
Property 2 :  (m, n)  2  sin 2 m 1  cos 2 n1  d
0

1
We have  (m, n)  x (1  x) dx 
m1 n 1
Proof :
0

Put x  sin
2


dx  2sin  cos d


When x  0 ,   0 and x  1 ,  
2
 /2
 ( m, n )   (sin  ) m 1 (1  sin 2  ) n 1 .2 sin  cos  d
2

 /2
 ( m, n )   sin  cos 2 n  2  2 sin  cos  d
2m2

 /2
 (m, n)  2  sin
2 m 1
 cos 2 n 1  d
0


1  p 1 q  1
Note : 0
2
sin p  cos q  d   
2  2
,
2 
 letting

2 m  1  p, 2 n  1  q

Property 3 :  (m, n  1)   (m  1, n) =  (m, n)

Proof : LHS =  (m, n  1)   (m  1, n)

1 1
x m 1
(1  x) dx   x m (1  x)n 1 dx
n

0 0

1

  x m1 (1  x) n  x m (1  x) n 1 dx 
0
1

  x m1 (1  x) n1 [(1  x)  x] dx 
0
1
  x m1 (1  x) n 1dx
0

  (m, n)
  m, n  1  ( m  1, n)  ( m, n )
Property 4 :  
n m mn

Proof : By definition
1
 (m, n  1)   x m1 (1  x)n dx
0

Integrating by parts

1 1
xm xm
 (1  x) n
 n(1  x) n 1 (1)dx
m 0 0
m
1
n
 0
m0 x m (1  x) n 1 dx

n
  (m  1, n)
m

a c a c ac
we know that, if  then  
b d b d bd

 ( m, n  1)  ( m  1, n )  ( m  1, n )   ( m, n  1)  ( m, n )
   
n m mn mn

t m 1
Property 5 :  (m, n)   mn
dt
0 (1  t )

Proof : We have
1
 (m, n)   x m 1 (1  x) n1 dx
0

t x
put x  t 
t 1 1 x
t
and dx  dt
(t  1) 2

when x0 t 0

x 1 t 
 m 1 n 1
 t   t  1
  ( m, n )     1   . dt
0
1 t   1 t  (1  t ) 2
  
t m 1 1 1 t m 1
 . . dt  0 (1  t )m 1 dt
0
(1  t ) m 1 (1  t ) n 1 (1  t ) 2
Relation between Beta and gamma function:
 ( m ) ( n )
1.  ( m, n ) 
( m  n)
Proof:

2
 (m, n)  2  sin 2 m 1  cos 2 n 1 d ......(1)
0

(n)   x n 1e  x dx.......( 2)
0

put x  t 2  dx  2tdt
and x  0  t  0, xt 

Now (2) becomes,


( n)   t 2   n 1
e t 2tdt
2

0

(n)  2  t 2 n 1e t dt.......(3)
2

Consider ,

(n)  2  x 2 n 1e  x dx.........( 4)
2

0

(m)  2  y 2 m 1e  y dy........(5)
2

0

(m  n)  2  r 2 ( m  n ) 1e  r dr.......(6)
2

0

(m)(n)  4   x 2 n 1 y 2 m1e ( x
2
 y2 )
dxdy
0 0

Let us evaluate RHS by changing into polar coordinates using x  r cos  , y  r sin 
We have x2+y2=r2

Also, dxdy  rdrd


And  varies 0 from and r varies 0 from  .
2
 
 2
(m)(n)  4  e  r (r cos  ) 2 n 1 (r sin  ) 2 m 1rdrd
2

0 0


2 
 r 2 2 n 1 2 m 1  
 4 e r r   sin 2 m 1
 cos  d dr
2 n 1

0 0 
 

 2 
 
 2 e r dr 2  sin
2
r 2[( m  n ) 1] 2 m 1
 cos  d ..........(7)
2 n 1

0  0 
 
comparing (1), (6) and (7)
(m)(n)  (m  n)  (m, n)
 ( m) ( n )
 (m, n) 
 (m  n)
Some Useful results:
1
Result 1 :    
2
1 1
   
1 1 2 2
Proof : We have   ,      
2 2  1 1 
  
2 2
 2
1
 
1 1 2
  ,     .......(1)
2 2 (1)
Now consider

2
 m, n   2  sin 2 m 1  cos 2 n 1 d
0

2
1 1
  ,   2  sin 0  cos 0 d
2 2 0

2
 2  1d
0

 2 x 02
1 1
  ,    ........( 2)
2 2
Put the value of (2) in (1)
  1 2 
   
   2  
1
1
   
2
Result 2 : Duplication formula

 1 
(m)  m    2 m 1 (2m)
 2 2

 (m) (n)
Proof : We have  (m, n) 
(m  m)
 
2
 ( m ) ( n )
 (m, n)  2  sin 2 m1  cos 2 n1 d  ..........(1)
0
 ( m  n)
1
put n  in (1)
2
 1
2
 ( m)  
 1   2  .......(2)
  m,   2  sin 2 m1  d 
 2   1
0
 m  
 2
put m  n in (1)

2
((m)) 2
 (m, m )  2  sin 2 m1  cos 2 m1  d 
0
2m 
consider ,

2
 2  sin  cos  
2 m 1
d
0

2 m 1
 sin 2 
2
 2   d
0
2 

2
1
 sin 2 
2 m 1
2 2 m 1
d
2 0

d
put 2    d  ,  var ies from 0 to 
2


1 2
d
 sin 
2 m 1
2 2 m 1
2 0
2

2
1
 sin d
2 m 1

2 2 m1 0


 2

 
Using the fact that, sin k d  2 sin k d
0 0
 
2
1
 2 m 1
2  sin 2 m1d
2 0

2
((m)) 2
2 2 m 1  (m, m)  2  sin 2 m 1d  2 2 m 1 ..........(3)
0
 ( 2 m)
with reference to (3) and (2) we can say that ,
 
2 2
2  sin 2 m1d  2  sin 2 m 1d
0 0

since the variable is arbitrary in a definite integral

1
 ( m ) 
 2   2 2 m1 ((m))
2

 1  ( 2m )
 m  
 2
 ( m)  ((m)) 2
 2 2 m 1
 1  ( 2 m)
 m  
 2
 1
 (2m)  2 2 m 1 (m) m  
 2
and
 1
  m,   2 2 m1  m, m 
 2

Problems:



2
2 n 1  at
Prove that (n)  2a t e dt
n
1.
0

Solution: By definition  ( n)   e  x x n 1dx
0

Put x  at 2
 dx  2atdt

(n)   e  at (at 2 ) n1 (2atdt )
2

0

(n)  2a n  t 2 n 1e at dt
2

0
 


2
4 x
2. Evaluate x e dx
0

I   x 4 e  x dx
2
Solution :
0

dt
Put t  x 2 , dx 
2 t

dt
  t 2e  t
0 2 t

1  3 / 2 t
  t e dt
20


1 1 5 1  3  1 3  3
  t 5/21e t dt         1   
20 2 2 2  2  2 2  2

1 3 3 1 3 1  1 3 1 1
 . .     . .    1  . .   
2 2 2 2 2 2  2 2 2 2

3
 
8

dx
3. Evaluate  2

0 34 x

1
Solution: I  3
0
4 x2
dx

Put 3  em , 4x2  t

t 1
x  dx  t 1/ 2 dt
2 4

1 1/ 2
t dt
 4
0 e mt

 1
1  2 mt
4 0
 t e dt

Put mt  u
 mdt  du

1 / 2
1  u  du 1  1
1
 .   e u
 u 2
e u
du
4 0m m 4 m 0

1 1
  
4 m  2

1
 
4 log 3

1
4. Evaluate  ( x log x) 4 dx
0
1
Solution: I   ( x log x ) 4 dx
0

log x  t
 x  e t
dx  e  t dt

x  0  t   and x  1  t  0
0
  e 4t  t  e  t dt
4


  t 4 e 5t dt
0

du
Put 5t = u  dt 
5
4
1  u  1 
    e u du  5  u 51e u du
5 0 5 5 0

1
  (5)
55

4!

55
 9  7
5. Evaluate: i)    ii)    
 2  2
9 7 
Solution: i)        1
2 2 
7 7 7 5  7 5 5 7 5 3  3
        1  .     . .   
2  2 2  2  2 2  2 2 2 2  2 
7 5 3 3 1
 . . .  
2 2 2 2 2
7 5 3 1 105
 . . .   
2 2 2 2 16

 7   5 
    1     1
 7 2  2  5 2
   
2 
ii)       . 
 2 7 7  2 7 5
 
2 2

 3 
    1

4  3
   
4  2   8   1 
 
35  2  35 3 105  2 

2

 1 
  1
8  2 

105 1
2
16  1 
  
105  2 
16
 
105
 /2
6. Evaluate 
0
tan  d .

 /2
Solution: I   tan  d 
0
 /2
  sin
1/ 2
 cos 1/ 2  d
0

1  1 / 2  1 1 / 2  1 
  , 
2  2 2 
1 3 1
   , 
2 4 4
 3 1
     
1 4 4

2  (1)
 1 1
 1     
1  4  4 
 Use  ( n ) (1  n ) 
2 1 sin n
1 

2  
sin  
4


2
  2
e y
 dy 
 y2
7. Evaluate ye dy
0 0 y
 1

y
2
Solution: I1  2
e  y dy
0
1
y2  u  y  u 2
1 12 1
 dy  u du
2
 1 1
1 2 1
I1   u 4 e  u . u du
0 2
3
1  4 1 u
2 0
 u e du

1  3
I1     ………(1)
2  4
 2
e y
I2   dy
0 y
Put y 2  u
1
 y  u2
1 12 1
 dy  u du
2
 1 1
1 1
I2   e  u u 4 . u 2
0
2
 1
1 1
  e  u u 4 du
20
1 1
I2     …………( 2 )
2  4
 1  3 1 1
   
I1 . I 2 
2  4 2 4
1 1  1
     1  
4 4  4
1   
 .  
4 sin  4 1 2 2
4 2
 m 1 
x x n1
8. Prove that  (m, n)   dx  0 (1  x)mn dx
0
(1  x) m n
1


We have  (m, n)  x (1  x) dx
m 1 n 1
Solution:
0

1
Put x 
1 y
1
 dx  dy
(1  y)2
1 1 x
Now y  1  y 
x x
When x  0  y   and x  1  y  0
0 m 1 n 1
 1   1  1
 ( m, n )     1   dy

1 y   1 y  (1  y ) 2

y n1
 m n
dy
0 (1  y )
Replace y by x,

x n 1
 ( m, n )   mn
dx
0 (1  x )

Since  (m, n)   (n, m)


x m 1
  (1  x) m  n dx
0
2
9. Evaluate  x (8  x 3 )1/3 dx
0
2
Solution: I   x (8  x )
3 1/3
dx
0

Put x3  8t
 x  2t1/3
1 13 1
 dt  2 t dt
3
When x  0  t  0 and x  2  t  1
 1 1 1 1
2 1
  2t (8  8t ) t 3 dt
3 3

0
3
2
4 1 3 1 1
  t . 2 1  t 3 dt
30
2
8 1 3 1 4
  t 1  t  3
1
dt
30
2 4  21 1
        
8 2 4 8  3  3 8  33 3
  ,   
3 3 3 3  ( 2) 3 1
8  1 1
   1    
9  3 3
8  
 Use  ( n ) (1  n ) 
9 sin  sin n
3
8  16
 . 
9 3 9 3
2
b
10. 
Show that ( x  a ) m (b  x) n dx  (b  a) m  n 1  (m  1, n  1) where ‘a’ and ‘b’ are
a
unequal constants.
b
Solution: Let I   ( x  a ) m (b  a ) n dx
a

Put x - a  t (b - a)

x  a  t  0 and x  b  t  1
1
 I   t m (b  a ) m b  a  t (b  a ) (b  a ) dt
n

1
  t m (b  a) m (b  a) n 1  t  (b  a) dt
n

1
 (b  a ) m  n 1  t m (1  t ) n dt
0

 (b  a)mn1  (m 1, n 1) .

Tracing of standard curves:

Introduction

In many engineering applications we need the length of a curve or something distributed

along the length of a curve. Area of some region or something over the region bounded by closed
curve, volume of solids formed by the revolution of the area bounded by a curve about a line
something distributed in that volume, surfaces formed by the revolution of the arc of a curve about the
line.

Procedure for tracing Cartesian Curves

Let f ( x , y )  0 represent the equation of the given Cartesian curve.

Symmetry of the curve

A curve is symmetrical about x -axis, if f ( x ,  y )  f ( x , y) or y occurs only in even powers.

A curve is symmetrical about y -axis, if f ( x , y)  f ( x , y) or x occurs only in even powers.

A curve is symmetrical about the origin if f ( x ,  y )  f ( x , y) or both x and y occurs in even

powers.

A curve is symmetrical about the line y  x if on interchanging x and y , the equation of the curve
remains unchanged.

Origin

If x  0, y  0 satisfy the equation of the curve, then the curve passes through origin

If it does, then find the equation of the tangents at the origin by equating the lowest degree terms in
the equation to 0 .

Asymptotes

See if the curve has only asymptote parallel to axis.

Asymptotes Parallel to y axis

Equate the co-efficient of highest degree term in y to 0 , to obtain asymptotes parallel to y -axis.

Asymptotes Parallel to x axis

Equate the co-efficient of highest degree term in x to 0 , to obtain asymptotes parallel to x -axis.

Oblique Asymptotes

Asymptotes not parallel to the axis, called oblique asymptotes, are determined by putting y  mx  c
in the equation of the curve and finding both possible values of m and c by equating the coefficients
of the first two highest powers of x to 0 .
Points of Intersection

The points of intersection of the curve with the x -axis and the y - axis are found by putting x  0 and
y  0 respectively and solving the equation of the curve.

Region of existence

Find the region of existence of the curve by considering all the above points.

Problems

1. Trace the curve Leminscate given by the equation ( x  y )  a ( x  y )

2 2 2 2 2 2

Solution:

Symmetry: Both x and y occur only in even powers, therefore the curve is symmetrical about both the
coordinate axis.

Origin: The point 0 , 0  satisfies the equation and hence the curve passes through the origin.

Tangents at the origin: The tangents at the origin are found by rewriting the given equation in the
form

 
x 4  y 4  2x 2 y 2  a 2 y 2  x 2  0

and equating the lowest degree term in this equation to zero.

This gives us the equation

y2  x2  0  y  x

Hence y  x and y   x are the tangents at the origin.

Points of Intersection: For y  0 , the equation gives x  0,  a , hence the points of intersection are
 a , 0 
Asymptotes: Asymptotes does not exist.

Region of existence:

The curve lies in the region for which x 2  a 2 i.e. between x  a and x  a .
 2. Trace the curve Strophoid y 2 (a  x)  x 2 (a  x) .
Solution:

Summetry: The equation contains only even powers of y . Thus the curve is symmetrical about x -
axis.

Origin:The point 0 , 0  satisfies the equation and hence the curve passes through the origin
origin.

Tangents at the origin:Equating

Equating the lowest degree terms to zero, we have th
the equation of tangents at
the origin as
y 2  x 2 (i.e., y   x ).
The two tangents are real and district.
Points of Intersection:When y  0 , we have x  0 and x   a .

axis only at  0,0  and  0, a  .Further the curve meets the y -axis only at
The curve meet the x -axis

 0, 0  .
Asymptotes: Equating the highest degree term in y , we get the asymptote parallel to y -axis as
a  x  0 (i.e., x  a is the only asymptote which is a line parallel to the y-axis).

Region of existance:

x 2 (a  x)
Now, y (a  x)  x (a  x)  y  
2 2

(a  x )

x 2 (a  x)
Taking +ve sign, we have y  
(a  x )

When x  a or x   a , y is imaginary.

 The curve does not lie beyond the lines.

i.e., the curve lies between the lines x   a and x  a .

Procedure for tracing Polar Curves:

Let f (r , )  0 represent the equation of the given Cartesian curve

Symmetry

A curve is symmetrical about the initial line, if the equation remains unchanged when  is changed to


A curve is symmetrical about the line through the pole perpendicular to the initial, if the equation
remains unchanged when  is changed to    )

A curve is symmetrical about the pole if only even powers of r occur in the equation(i.e., it remains
unchanged when r is changed to  r )

Limits

Determine numerically the greatest value of r so as to notice whether the curve lies within a circle or
not

Determine the region in which no portion of the curve lies by finding those values of  for which r
is imaginary

Points of Intersection

Giving successive values to  , find the corresponding values of r

Determine the points where the tangent coincides with the radius vector or is perpendicular to it. (i.e.,
d
the point where tan   r   or  ).
dr

3. Trace the curve Cardioid given by the equation r  a (1  cos  ) .

Solution:

Summetry: When 𝜃is changed to−𝜃, the equation remains unaltered, therefore the curve is
symmetrical about the initial line.

Limits: Since | cos  |  1, we have r  2 a , therefore the entire curve lies within the circle centered
at the pole with r  2 a as the radius.

When    , r  0 , therefore the curve passes through the pole and the tangent is there at the line
  .

dr
Points of Intersection : Here   a sin  ,
d

d a(1  cos )     
so that tan  r    cot   tan  
dr  a sin 2  2 2
    0 , therefore at the point   0 (for which r  2a ) on the curve,
  at
2

the tangent is perpendicular to the radius vector.

r  a When    / 2 , therefore the curves cuts the line    / 2 at a ,  / 2  .

As  increases from 0 to  , then r decreases from 2a to a and as  increases from  / 2 to  , further

decreases from a to 0.

As set of corresponding values of  and r on the curve are shown below:

 0  /2 2 / 3  3 / 2 2

r 2a a a/2 0 a 2a

Procedure for tracing Parametric Curves:

Let the equation of the curve to be traced is given by the equation x  f (t ) and y   (t )

Symmetry

A curve is symmetrical about the x -axis if on replacing t by  t , f (t ) remains unchanged and  (t )

changes to   (t )
A curve is symmetrical about the y  axis if on replacing t by  t , f (t ) changes to  f (t ) and  (t )
remains unchanged
A curve is symmetrical in the opposite quadrant if on replacing t by  t , both f (t ) and remains
unchanged.
Limits

Find the greatest and the least values of and y so as to determine the strips parallel to the axes within
or outside which the curve lies.

Asymptote

If the value of a parameter makes either x or y infinite, then that value of t gives asymptotes parallel
to either the x or the y -axis respectively
Special Point

i) Determine the points where the curve crosses the axes. The points of intersection of the curve with
the x -axis are given by the roots of  (t )  0 , while those with the y  axis are given by the roots
of f (t )  0

ii) Giving t a series of values, plot the corresponding values of x and y , noting whether x and y
increase or decrease for the intermediate values of t . For this purpose, we consider the signs of
dx dy
and for different values of t .
dt dt

iii) Determine the points where the tangent is parallel or perpendicular to the x -axis (i.e.,where
dy
 0 or   )
dx

iv) When x andy are periodic functions of t with a common period, we need to study the curve only
for one period as the other values of t will repeat the same curve.

4. Trace the curve Astroid given by the equation x

2/3
 y2/3  a2/3 also given by the parametric
equations in the form x  a cos 3 t , y  a sin 3 t

Solution:

Symmetry: curve is symmetrical about both the coordinates axes and also about the lines y   x

Origin: The curve does not pass through origin and hence no tangents at the origin

points of intersection: For t  0 and t   , we have x  a and y  0 therefore  a , 0  are the

points of intersection.

Similarly, for t   / 2 and t  3 / 2 , we have x  0 and y   a , therefore 0 ,  a  are the points of

intersection on the axis of reference

Since | x |  a | cos t |3  a and | y |  a | sin t |3  a ,

the entire curve lies in the region for which  a  x  a,  a  y  a

dy dy / dt 3a sin 2 t cos t
We have y '      tan t
dx d / dt  3a cos 2 t sin t

At the points (a, 0) , we have y  0 , therefore x  axis is the tangent at the points (a, 0) and
similarly, y  axis is the tangent the points (0,  a) .

In the first quadrant (where x  0 , y  0 ) we have tan t  0 , therefore y  0 and hence the
curve is decreasing in this quadrant.
Applications: Area under a plane curve, length of a plane curve, illustrative examples on
volume of revolution and surface area of revolution by a given curve (without proof).

Surface Area of Solid of Revolution:

Consider the revolution of an arc of a curve y  f (x) about the x  axis between the ordinates
x  a and x  b .

The Surface area of the solid so generated is called the volume of revolution generated by the curve.
This volume is given by the formula

x b 2
 dy 
S   2y 1    dx
x a  dx 

 If the axis of revolution is the y  axis, then the surface area is given by the formula

y d 2
 dx 
S   2x 1    dy Where y  c and y  d are appropriate limits of integration
y c  dy 

 In the case of polar curves, the volume generated by revolving the curve about the initial line
  0 x  axis) is given by

  2
 dr 
S   2r sin  r    d
2

   d 


 If the line of revolution is   ( y  axis), then the volume generated by revolving the
2
curve is given by
   2
 dr 
S 
 
2r cos r 2    d
 d 

Problems

1. Determine the surface area of the solid obtained by rotating y  9  x2 ,  2  x  2

about the x -axis.
x b 2
 dy 
Solution: The formula that we’ll be using here is, S   2y 1    dx
x a  dx 

dy x
Here 
dx 9  x2

2 2
 dy  x 3
 1    1 
 dx  9 x 2
9  x2
x 2
3
S  
x  2
2 9  x 2
9  x2
dx

x2
 
x 2
6 dx  24

2. Determine the surface area of the solid obtained by rotating the following parametric

curve about axis x  cos 3  , y  sin 3  , 0   
the x-axis .
2
Solution:

We’ll first need the derivatives of the parametric equations.

dx dy
 3 cos 2  sin  ,  3 sin 2  cos 
dt dt

  2 2
 dx   dy 
But S   2x      d
   d   d 

  / 2
S  2 cos 3   3 cos 2
 sin    3 sin 2  cos   d
2 2

 0

  / 2
  2 cos  3 cos  sin  d
3

 0
   / 2
 6  sin  cos  d
4

 0

 6 / 5

3. Find the area of the solid generated by the rotating of the loop of the curve
r 2  a 2 cos 2 about the initial line.
Solution:

The equation of the curve is r 2  a 2 cos 2

There are two loops of the curve. Differentiating the above equation we get

dr dr  a 2 sin 2
2r  2a sin 2 
2

d d r
  / 4  
 2a cos 2 sin  a 2 cos 2  a sin 2
2 2
 d
S2   cos 2 
 0  
  / 4
 1 
 4a 2
  sin   d  4a  cos  
2
0
/4
 4a 2 1  .
 0  2

4. Find the area of the solid formed by the revolution of the cardioid r  a 1  cos   of
about the initial line.
Solution:

The equation of the curve is r  a 1  cos  

The curve is symmetric about the initial line. The upper portion of the curve  varies

from 0 to 
dr
  a sin 
d
 
 2a 1  cos   sin  a 2 1  cos    a 2 sin 2   d

2
S
0
 

 
 2a 2  2 cos  / 2  2 sin  / 2 cos  / 2 2 cos  / 2 d
2

 0

 
32  a 2
 16  a  cos  / 2 sin  / 2   d 
2 4

 0 5
Volume of Solid of Revolution::

Consider the revolution of an arc of a curve y  f (x) about the x  axis between the ordinates
x  a and x  b .

The volume of the solid so generated is called the volume of revolution generated by the curve. This
volume is given by the formula

x b
V   y
2
dx
xa

If the axis of revolution is the y  axis, then the volume is given by the formula

y d

V   x
2
dy
y c

Where y  c and y  d are appropriate limits of integration

In the case of polar curves, the volume generated by revolving the curve about the initial line   0 (
x  axis) is given by

2
V   3 r sin  d
3


If the line of revolution is   ( y  axis), then the volume generated by revolving the curve is
2
given by

2
V   r 3 cos  d
3

appropriate limits for integration with respect to  have to be taken.

In the above two formulae, approp
Problems:

1. Determine the volume of the solid of revolution when the arc of the curve y  xe
x

between x  0 and x  1 is revolved about the line y  0 ( x  axis).

Solution:

The required volume is

1
V   y 2 dx
0

1
   x 2 e 2 x dx )
0


 (e 2  1)
4

2. Find the volume of the spindle-shaped solid generated by revolving the Astroid
x2/3  y 2/3  a2/3 about the x-axis.
Solution:

The required volume is twice the volume of the solid generated by the revolution of
the arc of the curve in the first quadrant (for which x varies from 0 to a ), about the
x  axis.
a
The required volume is V  2  y 2 dx 
0


We have x  a cos t , y  a sin t as the parametric equations and t 
3 3
when
2
x  0 & t  0 when x  a , we get
0
V  2  (a sin 6 t )  ( 3a cos 2 t sin t dt )
2

/2

 /2
 16a 3  sin
7
t cos 2 tdt
0

6.4.2 1 32 3
 6a 3 . .  a
9.7.5 3 105
 3. Find the volume generated by revolving the Cardioid r  a 1  cos   about   0 .
Solution:

Since the given curve is symmetrical about the initial line   0 .

The solid obtained by revolving the upper part of the curve about the initial line is the
same as the solid obtained by revolving the whole curve.

For the upper part of the given Cardioid,  varies from 0 to  .

Therefore the required volume is

 
2 2
V    r 3 sin  d   a 3  (1  cos  ) 3 sin  d
0
3 3 0

2
2
  a 3  t 3 dt , where t  1  cos 
3 0

2 24 8
  a3.   a3 .
3 4 3